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Mathematical Reasoning and Aptitude Questions and Answers

81.

A, B, C, D, E and F are sitting around a round table. A is between E and F. E is opposite to D and C is not in either of the neighbouring seats of E. The person opposite to B is

Exam Year(s)

UGC NET - 2013,

Answer: D

The person opposite to B is F.

Number the six seats 1 to 6 round the table. Put A in seat 1 with E in seat 2 and F in seat 6, so A sits between them. D is opposite E, so D is in seat 5. B and C take seats 3 and 4. C may not be next to E, and seat 3 is next to E, so C is in seat 4 and B is in seat 3.

The seat opposite seat 3 is seat 6, which is F. The mirror arrangement, with E and F exchanged, gives the same result: the person opposite B is F.

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82.

The missing term in the series 2, 7, 24, 77, ?, 723 is

Exam Year(s)

UGC NET - 2013,

Answer: A

The missing term is 238.

Each term is three times the previous term plus an odd number that grows by 2:

\[ 2 \times 3 + 1 = 7, \quad 7 \times 3 + 3 = 24 \]

\[ 24 \times 3 + 5 = 77 \]

\[ 77 \times 3 + 7 = 238 \]

\[ 238 \times 3 + 9 = 723 \]

The last step reaches the given 723, which confirms 238. The other options do not lead to 723 by the same rule.

Checking by working back is useful: from 723, subtracting 9 and dividing by 3 gives 238, and the same step repeated from 238 gives 77, which agrees with the series.

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83.

In certain city the Taxi charges comprise of a fixed charge and the charge of the distance travelled. A person paid ₹ 156 for a journey of 16 km and another person paid ₹ 204 for the journey of 24 km. The amount paid by a passenger who has travelled 30 km is

Exam Year(s)

UGC NET - 2013,

Answer: B

The passenger who travelled 30 km paid Rs 240.

Let the fixed charge be f and the charge per km be r.

\[ f + 16r = 156, \quad f + 24r = 204 \]

\[ 8r = 48 \Rightarrow r = 6, \quad f = 156 - 96 = 60 \]

\[ \text{Fare for 30 km} = 60 + 30 \times 6 = 240 \]

So Rs 240 is the fare. The other options come from a wrong rate or a missing fixed charge.

This is an example of a linear equation in two unknowns, and any pair of journeys gives the fixed charge and the rate per kilometre.

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84.

In certain code HEALTH is written as KHDOWK. The code of NORTH will be

Exam Year(s)

UGC NET - 2013,

Answer: A

The code for NORTH is QRUWK.

In HEALTH to KHDOWK each letter moves three places forward: H to K, E to H, A to D, L to O, T to W and H to K.

Apply the same shift to NORTH: N to Q, O to R, R to U, T to W and H to K. This gives QRUWK.

The other options change the order of the letters or shift them by different amounts, so they do not follow the rule.

A quick check is that the last letter H becomes K in both words, which confirms the shift of three, and the wrong options would not keep K at the end.

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85.

Yadav, Arjun, Rajesh and Kamal play cricket. Rajesh and Kamal play tennis but do not play football. Yadav does not play Badminton but play golf. Arjun does not play tennis. Kamal sometimes plays Badminton. Who does play only cricket ?

Exam Year(s)

UGC NET - 2013,

Answer: B

The person who plays only cricket is Arjun.

All four play cricket. Rajesh and Kamal also play tennis, so neither plays only cricket. Yadav plays golf, so he plays more than cricket. Kamal sometimes plays badminton as well.

Arjun does not play tennis and nothing in the statements gives him any other game, so he is the one who plays only cricket.

This type of question is solved by listing for each person the games that the statements give, and then picking the one with no game beyond cricket.

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86.

In a post-office, stamps of three different denominations of ₹ 7, ₹ 8, ₹ 10 are available. The exact amount for which one cannot buy stamps is

Exam Year(s)

UGC NET - 2014,

Answer: A

The exact amount that cannot be bought is 19. With stamps of 7, 8 and 10 the total 19 would need some combination of these values. Subtracting 7 leaves 12, subtracting 8 leaves 11 and subtracting 10 leaves 9, and none of 12, 11 or 9 can be made from 7, 8 and 10, so 19 is impossible.

The other amounts can be made: 20 = 10 + 10, 23 = 7 + 8 + 8 and 29 = 7 + 7 + 7 + 8. So 19 is the only amount in the list that cannot be bought. This type of problem is solved by testing each amount against sums of the available denominations.

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87.

In certain coding method, the word QUESTION is encoded as DOMESTIC. In this coding, what is the code word for the word RESPONSE ?

Exam Year(s)

UGC NET - 2014,

Answer: C

The code word for RESPONSE is OMESICEM. Matching QUESTION with DOMESTIC letter by letter gives the code: Q to D, U to O, E to M, S to E, T to S, I to T, O to I and N to C.

Applying this to RESPONSE: E to M, S to E, O to I, N to C, S to E and E to M, while R and P are not in QUESTION and are read from the options as O and S. Only OMESICEM agrees with every letter we can check. OMESUCEM, OMESICSM and OMESISCM each break the E, S, O or N correspondences.

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88.

If the series 4, 5, 8, 13, 14, 17, 22, .... is continued in the same pattern, which one of the following is not a term of this series ?

Exam Year(s)

UGC NET - 2014,

Answer: C

The number that is not a term of the series is 33. The differences repeat in the cycle +1, +3, +5: 4, 5 (+1), 8 (+3), 13 (+5), 14 (+1), 17 (+3), 22 (+5).

Continuing the cycle gives 23 (+1), 26 (+3), 31 (+5), 32 (+1), 35 (+3), 40 (+5). So 31, 32 and 35 all appear in the series, while 33 never does, and 33 is the odd one out. A quick check of 33: after 32 the next step is +3, which lands on 35, so 33 is skipped.

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89.

Complete the series BB, FE, II, ML, PP, ...........by choosing one of the following option given :

Exam Year(s)

UGC NET - 2014,

Answer: A

The next pair is TS. Take the first letters B, F, I, M, P: the steps are +4, +3, +4, +3, so the next is +4, which gives T. Take the second letters B, E, I, L, P: the steps are +3, +4, +3, +4, so the next is +3, which gives S.

The pair is therefore TS. ST has the letters reversed, and RS and SR break the +4 step for the first letter, since R and S are only +2 and +3 from P. Writing the pairs out in order, BB, FE, II, ML, PP, TS, shows both alternating patterns running together.

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90.

A man started walking from his house towards south. After walking 6 km, he turned to his left and walked 5 km. Then he walked further 3 km after turning left. He then turned to his left and continued his walk for 9 km. How far is he away from his house ?

Exam Year(s)

UGC NET - 2014,

Answer: C

He is 5 km from his house. Take the start as the origin. Walking 6 km south puts him at (0, -6). Turning left from south means facing east, so 5 km takes him to (5, -6). Turning left again he faces north and walks 3 km to (5, -3). Turning left again he faces west and walks 9 km to (-4, -3).

\[ d = \sqrt{4^{2} + 3^{2}} \]

\[ d = \sqrt{25} = 5 \text{ km} \]

The distances 3, 4 and 6 do not match this result. The straight line from the start to the end point is the hypotenuse of a 3-4-5 triangle.

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Tags: Mathematical Reasoning and Aptitude Questions and Answers